A circle with center $O$ has radius $8$ units and circle $P$ has radius $2$ units.  The circles are externally tangent to each other at point $Q$.  Segment $TS$ is the common external tangent to circle $O$ and circle $P$ at points $T$ and $S$, respectively.  What is the length of segment $OS$?  Express your answer in simplest radical form.
Answer: We create a diagram with the given information from the problem: [asy]
draw(Circle((0,0),8));
draw(Circle((10,0),2));
dot((0,0));dot((10,0));
label("$O$",(0,0),SW); label("$P$",(10,0),SW);

dot((8,0)); label("$Q$",(8,0),SW);

label("$T$",(4.6,6.6),NE); label("$S$",(11,1.7),NE);
draw((4.6,6.6)--(11,1.7));
[/asy]

We draw in radii $OT$ and $PS$ and connect $O$ and $P$.  Then we drop a perpendicular from $P$ to $OT$ that intersects $OT$ at $R$:

[asy]
draw((0,0)--(4.6,6.6),red); draw((10,0)--(11,1.7),blue);
draw(Circle((0,0),8));
draw(Circle((10,0),2));
dot((0,0));dot((10,0));
label("$O$",(0,0),SW); label("$P$",(10,0),SW);

label("$T$",(4.6,6.6),NE); label("$S$",(11,1.7),NE);
draw((4.6,6.6)--(11,1.7));
draw((0,0)--(8,0),red); draw((8,0)--(10,0),blue);
draw((10,0)--(3.3,4.8));
label("$R$",(3.3,4.8),W);
[/asy]

$\angle OTS$ and $\angle PST$ are right angles as tangents create right angles with radii at points of tangency.  $RTSP$ is a rectangle, and $\triangle ORP$ is right.  We use Pythagorean theorem on $\triangle ORP$:  we have $OP=8+2=10$ and $OR=8-2=6$, so $RP=\sqrt{OP^2-OR^2}=\sqrt{10^2-6^2}=8$.  Then $TS=8$ as well.

[asy]
draw((0,0)--(4.6,6.6));

label("8",(2,3),N); label("8",(8,5));
draw(Circle((0,0),8));
draw(Circle((10,0),2));
dot((0,0));dot((10,0));
label("$O$",(0,0),SW); label("$P$",(10,0),SW);

label("$T$",(4.6,6.6),NE); label("$S$",(11,1.7),NE);
draw((4.6,6.6)--(11,1.7));
draw((0,0)--(11,1.7));
draw((10,0)--(11,1.7));
[/asy] Finally, $OS$ is the hypotenuse of right triangle $\triangle OTS$ with $OT=TS=8$.  Hence $OS=\sqrt{8^2+8^2}=\boxed{8\sqrt{2}}$.